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3 years ago

Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc

Physics

2 answers:

motikmotik3 years ago

4 0

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

$E = mc^2$

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

$E = mc^2$

const2013 [10]3 years ago

4 0

the answer is B i just took it

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A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

so:

Mgh = $\frac{1}{2}IW^2 +\frac{1}{2}MV^2$

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = $\frac{1}{2}MR^2$

I = $\frac{1}{2}(5kg)(2m)^2$

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = $\frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2$

Replacing the data:

(5kg)(9.8)(0.3m) = $\frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2$

solving for V:

(5kg)(9.8)(0.3m) = $V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))$

V = 2 m/s

8 0

3 years ago

Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car i

Korvikt [17]

As per law of Heisenberg uncertainty law

product of uncertainty in position and uncertainty in momentum will be constant

$\Delta x . \Delta P = \frac{h}{4\pi}$

$\Delta x . m \Delta v = \frac{h}{4\pi}$

now plug in all data

$(5\times 0.3048). (2250 \times 0.454) \Delta v = \frac{6.6 \times 10^{-34}}{4\pi}$

$\Delta v = 3.37 \times 10^{-38} m/s$

So above is the uncertainty in velocity of the object

4 0

2 years ago

Read 2 more answers

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a

seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

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3 years ago

A car travels 92 miles in 2 hours. What is the car's AVERAGE SPEED?

nekit [7.7K]

Answer:

about 46 mph

Explanation:

6 0

2 years ago

Can anyone heelp me plzz

lina2011 [118]

Answer:

the ans is D... good luck

3 0

2 years ago